Question

A jet pilot takes his aircraft into a vertical loop. If the jet is moving at a speed of 700km/hr at the lowest point in the loop, determine the radius of the circle so that the acceleration which the pilot experiences at the lowest point of the loop does not exceed 6 time the acceleration due to gravity or (6g’s).

Answer 1

Answer:

The radius of circle is 643 m.

Explanation:

The centripetal acceleration of the jet is

$a=\frac{V^{2}}{r}$

Where,

a is "centripetal acceleration"

V is "speed "

r is "radius of the circle "

Given values ,

Speed of the jet (v) = 700 km/h

(1 km = 1000 m; 1 h = 3600 sec. To convert km/h into m/sec, multiply the number by 5 and then divide it by 18.)

Speed of the jet (v)$=\left(700 \times \frac{5}{18}\right) \mathrm{m} / \mathrm{s}$

V = 194.444 m/s

a = 6g's

$\text { (g is referred to as the acceleration of gravity. Its value is }\left.9.8 \mathrm{m} / \mathrm{s}^{2} \text { on Earth }\right)$

$a=6 \times 9.8 \mathrm{m} / \mathrm{s}^{2}$

$a=58.8 \mathrm{m} / \mathrm{s}^{2}$

$\text { The radius of circle is a }=\frac{V^{2}}{r}$

Substitute the given values in the formula,

$a=\frac{194.444^{2}}{58.8}$

$a=\frac{37808.46}{58.8}$

a = 643 m

The radius of circle is 643 m.